这不是作业问题。在一次面试中，这个问题被问给了我的一个 friend 。
我有一个从文件中读取的行的list作为输入。每行的开头都有一个标识符，例如(A，B，NN，C，DD)。根据标识符，我需要将记录列表映射到包含对象层次结构的单个对象A中。
层次结构描述:
每个A可以具有零种或多种B类型。
每个B标识符可以将零个或多个NN和C作为子代。类似地，每个C段可以具有零个或多个NN和DD子级。 Abd每个DD可以有零个或多个NN作为子代。
映射类及其层次结构:
所有类都将具有value来保存当前行中的String值。
**A - will have list of B**
class A {
List<B> bList;
String value;
public A(String value) {
this.value = value;
}
public void addB(B b) {
if (bList == null) {
bList = new ArrayList<B>();
}
bList.add(b);
}
}
**B - will have list of NN and list of C**
class B {
List<C> cList;
List<NN> nnList;
String value;
public B(String value) {
this.value = value;
}
public void addNN(NN nn) {
if (nnList == null) {
nnList = new ArrayList<NN>();
}
nnList.add(nn);
}
public void addC(C c) {
if (cList == null) {
cList = new ArrayList<C>();
}
cList.add(c);
}
}
**C - will have list of DDs and NNs**
class C {
List<DD> ddList;
List<NN> nnList;
String value;
public C(String value) {
this.value = value;
}
public void addDD(DD dd) {
if (ddList == null) {
ddList = new ArrayList<DD>();
}
ddList.add(dd);
}
public void addNN(NN nn) {
if (nnList == null) {
nnList = new ArrayList<NN>();
}
nnList.add(nn);
}
}
**DD - will have list of NNs**
class DD {
String value;
List<NN> nnList;
public DD(String value) {
this.value = value;
}
public void addNN(NN nn) {
if (nnList == null) {
nnList = new ArrayList<NN>();
}
nnList.add(nn);
}
}
**NN- will hold the line only**
class NN {
String value;
public NN(String value) {
this.value = value;
}
}
我到目前为止所做的事情:
方法
public A parse(List<String> lines)读取输入列表，并返回对象A。由于可能存在多个B，因此我创建了单独的方法'parseB来解析每次出现的情况。在 parseB 方法中，循环遍历
i = startIndex + 1 to i < lines.size()并检查行的开头。出现“NN”被添加到B的当前对象中。如果在启动时检测到“C”，它将调用另一个方法parseC。当我们在开始时检测到“B”或“A”时，循环将中断。parseC_DD中使用了类似的逻辑。
public class GTTest {
public A parse(List<String> lines) {
A a;
for (int i = 0; i < lines.size(); i++) {
String curLine = lines.get(i);
if (curLine.startsWith("A")) {
a = new A(curLine);
continue;
}
if (curLine.startsWith("B")) {
i = parseB(lines, i); // returns index i to skip all the lines that are read inside parseB(...)
continue;
}
}
return a; // return mapped object
}
private int parseB(List<String> lines, int startIndex) {
int i;
B b = new B(lines.get(startIndex));
for (i = startIndex + 1; i < lines.size(); i++) {
String curLine = lines.get(i);
if (curLine.startsWith("NN")) {
b.addNN(new NN(curLine));
continue;
}
if (curLine.startsWith("C")) {
i = parseC(b, lines, i);
continue;
}
a.addB(b);
if (curLine.startsWith("B") || curLine.startsWith("A")) { //ending condition
System.out.println("B A "+curLine);
--i;
break;
}
}
return i; // return nextIndex to read
}
private int parseC(B b, List<String> lines, int startIndex) {
int i;
C c = new C(lines.get(startIndex));
for (i = startIndex + 1; i < lines.size(); i++) {
String curLine = lines.get(i);
if (curLine.startsWith("NN")) {
c.addNN(new NN(curLine));
continue;
}
if (curLine.startsWith("DD")) {
i = parseC_DD(c, lines, i);
continue;
}
b.addC(c);
if (curLine.startsWith("C") || curLine.startsWith("A") || curLine.startsWith("B")) {
System.out.println("C A B "+curLine);
--i;
break;
}
}
return i;//return next index
}
private int parseC_DD(C c, List<String> lines, int startIndex) {
int i;
DD d = new DD(lines.get(startIndex));
c.addDD(d);
for (i = startIndex; i < lines.size(); i++) {
String curLine = lines.get(i);
if (curLine.startsWith("NN")) {
d.addNN(new NN(curLine));
continue;
}
if (curLine.startsWith("DD")) {
d=new DD(curLine);
continue;
}
c.addDD(d);
if (curLine.startsWith("NN") || curLine.startsWith("C") || curLine.startsWith("A") || curLine.startsWith("B")) {
System.out.println("NN C A B "+curLine);
--i;
break;
}
}
return i;//return next index
}
public static void main(String[] args) {
GTTest gt = new GTTest();
List<String> list = new ArrayList<String>();
list.add("A1");
list.add("B1");
list.add("NN1");
list.add("NN2");
list.add("C1");
list.add("NNXX");
list.add("DD1");
list.add("DD2");
list.add("NN3");
list.add("NN4");
list.add("DD3");
list.add("NN5");
list.add("B2");
list.add("NN6");
list.add("C2");
list.add("DD4");
list.add("DD5");
list.add("NN7");
list.add("NN8");
list.add("DD6");
list.add("NN7");
list.add("C3");
list.add("DD7");
list.add("DD8");
A a = gt.parse(list);
//show values of a
}
}
我的逻辑工作不正常。您还有其他方法可以解决吗？您对我的方式有什么建议/改进吗？
最佳答案
使用对象的层次结构:
public interface Node {
Node getParent();
Node getLastChild();
boolean addChild(Node n);
void setValue(String value);
Deque getChildren();
}
private static abstract class NodeBase implements Node {
...
abstract boolean canInsert(Node n);
public String toString() {
return value;
}
...
}
public static class A extends NodeBase {
boolean canInsert(Node n) {
return n instanceof B;
}
}
public static class B extends NodeBase {
boolean canInsert(Node n) {
return n instanceof NN || n instanceof C;
}
}
...
public static class NN extends NodeBase {
boolean canInsert(Node n) {
return false;
}
}
创建一个树类:
public class MyTree {
Node root;
Node lastInserted = null;
public void insert(String label) {
Node n = NodeFactory.create(label);
if (lastInserted == null) {
root = n;
lastInserted = n;
return;
}
Node current = lastInserted;
while (!current.addChild(n)) {
current = current.getParent();
if (current == null) {
throw new RuntimeException("Impossible to insert " + n);
}
}
lastInserted = n;
}
...
}
然后打印树:
public class MyTree {
...
public static void main(String[] args) {
List input;
...
MyTree tree = new MyTree();
for (String line : input) {
tree.insert(line);
}
tree.print();
}
public void print() {
printSubTree(root, "");
}
private static void printSubTree(Node root, String offset) {
Deque children = root.getChildren();
Iterator i = children.descendingIterator();
System.out.println(offset + root);
while (i.hasNext()) {
printSubTree(i.next(), offset + " ");
}
}
}
关于parsing - 将字符串列表映射到对象的层次结构，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/9914480/